If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the minimum value of $(x^2+y^2)^2$

Question

If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the maximum value of $(x^2+y^2)^2$

My attempt is as follows:

$$(x-y)^2\ge 0$$ $$x^2+y^2\ge 2xy$$ $$2(x^2+y^2)\ge x^2+y^2+2xy$$ \begin{equation} 2(x^2+y^2)\ge (x+y)^2\tag{1} \end{equation}

Solving the given equation:

$$x^2+2xy-y^2=6$$ $$(x+y)^2=2y^2+6$$

So putting the value of $(x+y)^2$ in equation $1$

$$2(x^2+y^2)\ge 2y^2+6$$ $$x^2+y^2\ge y^2+3$$ \begin{equation} x^2\ge 3\tag{2} \end{equation}

So $x\in \left(-\infty,-\sqrt{3}\right) \cup \left(\sqrt{3},\infty\right)$

But how to proceed from here?

Answer 1

Use $$2(x^2+y^2)^2-(x^2+2xy-y^2)^2=(x^2-2xy-y^2)^2\geq0.$$

Answer 2

Perform a parametrization of the form $$x = r \cos \theta, \quad y = r \sin \theta.$$ Then we seek to minimize $(x^2 + y^2)^2 = r^4$ subject to the constraint $$\begin{align*} 6 & = r^2 \cos^2 \theta + 2 r^2 \cos \theta \sin \theta - r^2 \sin^2 \theta \\ &= r^2 \left( \cos^2 \theta - \sin^2 \theta + 2\cos \theta \sin \theta\right) \\ &= r^2 \left( \cos 2\theta + \sin 2\theta \right). \end{align*}$$ Therefore, $$r^2 = \frac{6}{\cos 2\theta + \sin 2\theta},$$ and on $\theta \in [0, 2\pi)$, the LHS is minimized when the denominator is maximized. Rewriting this using an additional trigonometric identity, $$\cos 2\theta + \sin 2\theta = \sqrt{2} \left(\cos 2\theta \sin \frac{\pi}{4} + \sin 2\theta \cos \frac{\pi}{4}\right) = \sqrt{2} \sin \left( 2\theta + \frac{\pi}{4}\right).$$ So the maximum is attained whenever $2\theta + \frac{\pi}{4} = 2\pi k + \frac{\pi}{2}$ for some integer $k$; namely $$\theta \in \left\{\frac{\pi}{8}, \frac{9\pi}{8}\right\},$$ and the maximum value is $\sqrt{2}$; thus the minimum value is $$r^4 = \left( \frac{6}{\sqrt{2}} \right)^2 = 18.$$ It is a straightforward exercise to compute the set of $(x,y)$ values for which this minimum is attained.

Answer 3

Play with linear combinations of $x^2+y^2$ and the expression from the given equation. This way we find for $a>1$: $$a(x^2+y^2) - 6 = (a-1)x^2-2xy+(a+1)y^2=(\sqrt{a-1}x-\sqrt{a+1}y)^2+(2\sqrt{a^2-1}-2)xy$$ We get rid of the last term if we let $a=\sqrt 2$: $$\tag1\sqrt 2(x^2+y^2) -6 = \left(\sqrt{\sqrt 2-1}\,x-\sqrt{\sqrt 2+1}\,y\right)^2\ge 0.$$ This gives us the lower bound $x^2+y^2\ge 3\sqrt 2$ (and so $(x^2+y^2)^2\ge18$). Is the inequality $(1)$ sharp, i.e., can equality hold? Clearly, equality holds iff $x=\sqrt{\sqrt 2+1}\,t$ and $y=\sqrt{\sqrt 2-1}\,t$ for some $t$. But does there actually exist $t$ such that the original condition is met? We compute $$ x^2+2xy-y^2=(\sqrt2+1)t^2+2t^2+(\sqrt2-1)t^2=\sqrt 2t^2$$ and this is indeed $=6$ for suitable $t$ (namely $t=\sqrt{3\sqrt 2}$).

Answer 4

The function to be minimized is the fourth power of the distance from the origin, so will have its extrema at the same points at which the distance is also an extremum. The constraint is a hyperbola, so the minimum distance occurs at its vertices, therefore this problem is equivalent to computing the fourth power of a semiaxis length of this hyperbola.

One approach is to compute the eigenvalues of the matrix $\frac16\begin{bmatrix}1&1\\1&-1\end{bmatrix}$, which work out to be $\pm\frac1{3\sqrt2}$. We want the positive eigenvalue. It is the reciprocal of the square of the semi-transverse axis length, so the minimum value of $(x^2+y^2)^2$ on this hyperbola is $(3\sqrt2)^2=18$.

Answer 5

Minimize $z=(x^2+y^2)^2$ s.t. $x^2+2xy-y^2=6$.

The contour curve is a circle with the center at the origin: $x^2+y^2=\sqrt z$.

To minimize $z$ implies to find the smallest radius of the circle. The smallest circle must touch the hyperbola. The slope of tangent to the hyperbola at $(x_0,y_0)$ is: $$2x+2y+2xy'-2yy'=0 \Rightarrow y'=\frac{x_0+y_0}{y_0-x_0}$$ The normal line to the hyperbola at $(x_0,y_0)$ must pass through the origin: $$y=\frac{x_0-y_0}{x_0+y_0}x$$ The intersection point of the hyperbola and the normal line is $(x_0,y_0)$: $$\begin{cases}x_0^2+2x_0y_0-y_0^2=6 \\ y_0=\frac{x_0-y_0}{x_0+y_0}\cdot x_0\end{cases} \Rightarrow \begin{cases}x_0^2+2x_0y_0-y_0^2=6 \\ x_0^2-2x_0y_0-y_0^2=0\end{cases} \stackrel{+}\Rightarrow x_0=\sqrt{3+y_0^2} \Rightarrow \\ 3+y_0^2+2\sqrt{3+y_0^2}\cdot y_0-y_0^2=6 \Rightarrow y_0^4+3y_0^2-\frac94=0 \Rightarrow \\ y_0^2=\frac{-3+\sqrt{18}}{2},x_0^2=\frac{3+\sqrt{18}}{2} \Rightarrow \\ z(x_0,y_0)=(x_0^2+y_0^2)^2=18 \ \text{(min)}$$ Desmos graph for reference.